s- integral
endpoints are 0 and pi/2
i need to find the integral of sin^2 (2x) dx.
i know that the answer is pi/4, but im not sure how to get to it.
i know:
s sin^2(2x)dx= 1/2 [1-cos (4x)] dx, but then i'm confused.
The indefinite integral of
(1/2) [1-cos (4x)] is
(x/2) - (1/2)(1/4)(sin 4x)
Evaluate that at x = p/2 and 0 and take the difference
I get [(pi/4) -0] - {0 - 0) = pi/4
YOu got the hardest part, now expand your expression.
Hint: what is the derivative of 1/8 sin(4x)?
To find the integral of sin^2(2x) dx, you can start by using the identity sin^2(x) = (1/2)(1 - cos(2x)). This allows us to rewrite the integral as:
∫ sin^2(2x) dx = ∫ (1/2)(1 - cos(4x)) dx
Now, let's integrate the expression (1/2)(1 - cos(4x)). To do this, we can use the power rule of integration, which states that the integral of x^n dx, where n is a constant, is (1/(n+1))(x^(n+1)). In this case, our constant is 0, so we have:
∫ 1 dx = x
Since the constant term -cos(4x) does not contain x, we can simply integrate it as if it were a constant term. The integral of -cos(4x) with respect to x is:
∫ -cos(4x) dx = -(1/4)sin(4x)
Putting it all together, we have:
∫ sin^2(2x) dx = (1/2) ∫ (1 - cos(4x)) dx
= (1/2) [∫ 1 dx - ∫ cos(4x) dx]
= (1/2) [x - (1/4)sin(4x)] + C
Here, C represents the constant of integration. To evaluate the definite integral with endpoints 0 and π/2, we substitute these values into the expression and take the difference:
∫[0,π/2] sin^2(2x) dx = [(1/2)(π/2) - (1/4)sin(4(π/2))] - [(1/2)(0) - (1/4)sin(4(0))]
= (π/4) - (1/4)sin(2π)
= (π/4) - (1/4)(0) (since sin(2π) = 0)
= π/4
Therefore, the value of the definite integral ∫[0,π/2] sin^2(2x) dx is π/4.